Question
Find: $\int\sin^{-1}(2\text{x})\text{dx}.$

Answer

$\int\sin^{-1}(2\text{x})\text{dx}$
Using ILATE rule
$\text{x}\sin^{-1}2\text{x}-\int\frac{2\text{x}}{\sqrt{1-4\text{x}^2}}\text{dx}$
$\text{x}\sin^{-1}2\text{x}+\frac{1}{4}\int\frac{-8\text{x}}{\sqrt{1-4\text{x}^2}}\text{dx}$
Taking $1-4\text{x}^2=\text{t}$
$\Rightarrow-8\text{x}\times\text{dx}=\text{dt}$
$\text{x}\sin^{-1}2\text{x}+\frac{1}{4}\int\frac{\text{dt}}{\sqrt{\text{t}}}$
$\text{x}\sin^{-1}2\text{x}+\frac{1}{4}\frac{2\text{t}^{\frac{1}{2}}}{1}+\text{C}$
$=\text{x}\sin^{-1}2\text{x}+\frac{\text{t}^{\frac{1}{2}}}{2}+\text{C}$
$=\text{x}\sin^{-1}2\text{x}+\frac{\sqrt{1-4\text{x}^2}}{2}+\text{C}$

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